1. ## tan x>x+x^3/3

Prove that $\displaystyle \tan x > x+\frac {x^3}3$ if $\displaystyle 0<x<\frac {\pi}2$.

2. Originally Posted by james_bond
Prove that $\displaystyle \tan x > x+\frac {x^3}3$ if $\displaystyle 0<x<\frac {\pi}2$.
Using Taylor series with remainder

$\displaystyle \tan x = x + \frac{x^3}{3} + 8 \tan c ( 1 + \tan^2 c) (2 + 3 \tan^2c ) \frac{x^4}{4!}$

where $\displaystyle 0<c<\frac {\pi}2$. Since

$\displaystyle 8 \tan c ( 1 + \tan^2 c) (2 + 3 \tan^2c ) \frac{x^4}{4!} > 0$

for

where $\displaystyle 0<c, x<\frac {\pi}2$, the result follows.

3. Other solutions? (Without Taylor series as we haven't learned it and it is too obvious from it :P)

4. Hi

Let f(x) = tan x - x - x^3/3
Then f'(x) = tan²x - x² = (tan x -x)(tan x + x)

For 0 < x < pi/2 tan x > 0 and x > 0 therefore tan x + x > 0
The sign of f'(x) is the same as g(x) = tan x - x
g'(x) = tan²x > 0
g is increasing on [0,pi/2[
g(0) = 0 so g(x) > 0 on [0,pi/2[

Therefore f'(x) > 0 on [0,pi/2[
f(0) = 0
f(x) > 0 on [0,pi/2[